What is the derivative of 1/x e^x ?

The derivative of 1/x e^x is -(e^x}{x^2}+\frac{e^x)/x

What is the first derivative of 1/x e^x ?

The first derivative of 1/x e^x is -(e^x}{x^2}+\frac{e^x)/x

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\frac{d}{d x} (\frac{1}{x} e^{x})

- \frac{e ^{x}}{x ^{2}} + \frac{e ^{x}}{x}

Solution steps

\frac{d}{d x} (\frac{1}{x} e^{x})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = \frac{1}{x}, g = e^{x}

= \frac{d}{d x} (\frac{1}{x}) e^{x} + \frac{d}{d x} (e^{x}) \frac{1}{x}

\frac{d}{d x} (\frac{1}{x}) = - \frac{1}{x ^{2}}

\frac{d}{d x} (e^{x}) = e^{x}

= (- \frac{1}{x ^{2}}) e^{x} + e^{x} \frac{1}{x}

Simplify

(- \frac{1}{x ^{2}}) e^{x} + e^{x} \frac{1}{x} : - \frac{e ^{x}}{x ^{2}} + \frac{e ^{x}}{x}

= - \frac{e ^{x}}{x ^{2}} + \frac{e ^{x}}{x}

\frac{d y}{d x} = a \cdot y

e^{t a n (5 t)}

(e^{f (x)})^{'}

\frac{d}{d x} ((\frac{x ^{2} + 2 x}{x + 1})^{5})

\frac{d}{d x} (9 ln (cos (x)))

What is the derivative of 1/x e^x ?
The derivative of 1/x e^x is -(e^x}{x^2}+\frac{e^x)/x
What is the first derivative of 1/x e^x ?
The first derivative of 1/x e^x is -(e^x}{x^2}+\frac{e^x)/x